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a(n) is the least j >= 2 such that (n+1)^A338136(n) mod n^j is a perfect power > 1.
3

%I #11 Oct 17 2020 10:52:51

%S 4,2,2,2,2,3,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,

%T 3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,

%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2

%N a(n) is the least j >= 2 such that (n+1)^A338136(n) mod n^j is a perfect power > 1.

%F (n+1)^A338136(n) mod n^A338150(n) = A338151(n).

%e a(2) = 4 because A338136(2) = 6 and 3^6 mod 2^4 = 3^2.

%e a(13) = 2 because A338136(13) = 2 and 14^2 mod 13^2 = 3^3.

%p g:= proc(n) local k, x, j, F;

%p for k from 2 to n-2 do

%p x:= (n+1)^k;

%p for j from 2 to floor(k*log[n](n+1)) do

%p F:= ifactors(x mod (n^j))[2];

%p if igcd(op(map(t -> t[2], F))) > 1 then return j fi

%p od od

%p end proc:

%p g(2):= 4: g(3):= 2:

%p map(g, [$2..40]);

%Y Cf. A001597, A338136, A338151.

%K nonn

%O 2,1

%A _J. M. Bergot_ and _Robert Israel_, Oct 12 2020

%E More terms from _Jinyuan Wang_, Oct 17 2020